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$2^{(x^3)} = 3^{(x^2)}$

Solve for x

I'm pretty sure I use logs to solve this, but how? to what base? I'm kinda lost.. Thanks

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  • $\begingroup$ Do you mean $2^{(x^3)}$ or $(2^x)^{3}$? (And the same question for the other term.) $\endgroup$ – Ben Whitney Sep 16 '15 at 0:41
  • $\begingroup$ Your equation is ambiguous, I have edited it. Please check that what I have done is what you want. $\endgroup$ – David Sep 16 '15 at 0:41
  • $\begingroup$ The first one. Thanks $\endgroup$ – FoolishNumber Sep 16 '15 at 0:42
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Taking logs to any base, $$x^3\log2=x^2\log3\ ,$$ and therefore either $x=0$ or $x=(\log3)/(\log2)$. Note that by the "change of base" formula, the last expression is the same no matter what base you are using.

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  • $\begingroup$ Ooooooooooo. I understand now. I had a misconception on how to take logs to a base. :(. Thank you! $\endgroup$ – FoolishNumber Sep 16 '15 at 0:49
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Take natural log $$ 2^{x^3} = 3^{x^2} \implies x^3 \ln 2 = x^2 \ln 3 \implies x^2 \left( x\ln2 - \ln 3\right) = 0 \implies x_{1,2} = 0,\ x_3 = \frac {\ln 3}{\ln 2} = \log_2 3 $$

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  • $\begingroup$ Thank you! I had a misconception on how to change bases :(. $\endgroup$ – FoolishNumber Sep 16 '15 at 0:50
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Note the following facts:

  1. $\log(a^{(b)}) = b\log(a)$. This is the key here (assuming $\log(a)$ exists).

  2. When you raise a value to 0 power, you get a value of 1, so $x^0=5^0=1$

It is obvious that when $x=0$, the 2 sides would be equal.

Now the other possible solution, starting from:

$2^{(x^3)} = 3^{(x^2)}$

Apply the log function to both sides to get:

$\log(2^{(x^3)}) = \log(3^{(x^2)})$

$x^3\log(2) = x^2\log(3)$

assuming $x$ is not zero,

$x=\log(3)/\log(2)=1.58496250072$

Now an interesting point here would be why would get the same answer if you used another base for the log?

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